Book of Evil 树双向DFS
来源:互联网 发布:java math.random 编辑:程序博客网 时间:2024/06/11 02:08
Book of Evil
Paladin Manao caught the trail of the ancient Book of Evil in a swampy area. This area contains n settlements numbered from 1 to n. Moving through the swamp is very difficult, so people tramped exactly n - 1 paths. Each of these paths connects some pair of settlements and is bidirectional. Moreover, it is possible to reach any settlement from any other one by traversing one or several paths.
The distance between two settlements is the minimum number of paths that have to be crossed to get from one settlement to the other one. Manao knows that the Book of Evil has got a damage range d. This means that if the Book of Evil is located in some settlement, its damage (for example, emergence of ghosts and werewolves) affects other settlements at distance d or less from the settlement where the Book resides.
Manao has heard of m settlements affected by the Book of Evil. Their numbers are p1, p2, ..., pm. Note that the Book may be affecting other settlements as well, but this has not been detected yet. Manao wants to determine which settlements may contain the Book. Help him with this difficult task.
The first line contains three space-separated integers n, m and d (1 ≤ m ≤ n ≤ 100000; 0 ≤ d ≤ n - 1). The second line contains mdistinct space-separated integers p1, p2, ..., pm (1 ≤ pi ≤ n). Then n - 1 lines follow, each line describes a path made in the area. A path is described by a pair of space-separated integers ai and bi representing the ends of this path.
Print a single number — the number of settlements that may contain the Book of Evil. It is possible that Manao received some controversial information and there is no settlement that may contain the Book. In such case, print 0.
6 2 31 21 52 33 44 55 6
3
Sample 1. The damage range of the Book of Evil equals 3 and its effects have been noticed in settlements 1 and 2. Thus, it can be in settlements 3, 4 or 5.
#include <iostream>#include <cstdio>#include <cmath>#include <vector>#include <map>#include <set>#include <string>#include <cstring>#include <algorithm>#include <iomanip>#include <queue>#include <cstdlib>#include <ctime>#include <stack>#include <bitset>#include <fstream>typedef unsigned long long ull;#define mp make_pair#define pb push_backconst long double eps = 1e-9;const double pi = acos(-1.0);const long long inf = 1e18;using namespace std;int n, m, d;int f[ 100100 ], g[ 100100 ];vector< int > graf[ 100100 ];bool ok[ 100100 ];void dfs1( int v, int p ){ //cout << v << " " << p << endl; f[v] = -1; for ( int i = 0; i < graf[v].size(); i++ ) { int next = graf[v][i]; if ( next == p ) continue; dfs1( next, v ); f[v] = max( f[v], ( f[next] == -1 ? -1 : f[next] + 1 ) ); } if ( ok[v] ) f[v] = max( 0, f[v] );}void dfs2( int v, int p, int root ){ g[v] = root; vector< int > sons; sons.pb( ( root == -1 ? -1 : root + 1 ) ); if ( ok[v] ) sons.pb( 1 ); for ( int i = 0; i < graf[v].size(); i++ ) { int next = graf[v][i]; if ( next == p ) continue; sons.pb( ( f[next] == -1 ? -1 : f[next] + 2 ) ); } sort( sons.begin(), sons.end(), greater<int>() ); for ( int i = 0; i < graf[v].size(); i++ ) { int next = graf[v][i]; if ( next == p ) continue; int nroot = sons[1]; if ( ( f[next] == -1 ? -1 : f[next] + 2 ) != sons[0] ) nroot = sons[0]; dfs2( next, v, nroot ); }}int main (int argc, const char * argv[]){ //freopen("input.txt","r",stdin); //freopen("output.txt","w",stdout); scanf("%d%d%d", &n, &m, &d); for ( int i = 1; i <= m; i++ ) { int a; scanf("%d", &a); ok[a] = true; } for ( int i = 1; i < n; i++ ) { int a, b; scanf("%d%d", &a, &b); graf[a].pb(b); graf[b].pb(a); } dfs1( 1, -1 ); dfs2( 1, -1, -1 ); int ans = 0; for ( int i = 1; i <= n; i++ ) if ( max( f[i], g[i] ) <= d ) ans++; //for ( int i = 1; i <= n; i++ ) cout << i << " " << f[i] << " " << g[i] << endl; cout << ans << endl; return 0;}
- Book of Evil 树双向DFS
- CodeForces 337D Book of Evil(双向dfs)
- CF337D Book of Evil【dfs】
- Codeforces 337D Book of Evil 【树,dfs】
- codeforces 337E Book of Evil (dfs)
- 337D Book of Evil
- Codeforces 337D Book of Evil (树的直径)
- Codeforces 337 D Book of Evil(树形dp,两遍dfs)
- CF-337D Book of Evil
- codeforces 337D Book of Evil
- cordeforces 337D Book of Evil
- CodeForces Round #196 D Book of Evil
- codeforces 338b Book of Evil
- 【树形DP】 codeforces 337D Book of Evil
- CF 337D(Book of Evil-distdown[]-distup[])
- 树形dp-CF-337D. Book of Evil
- codeforces 337D D. Book of Evil(树形dp)
- CodeForces 337D——Book of Evil(数据结构)
- Objective-C中防止重复代码的一种方法
- Java中继承thread类与实现Runnable接口的区别
- codeforces 340 A. The Wall
- linux下如何添加一个用户并且让用户获得root权限
- jQuery之使用jQuery.fn.prop()替换jQuery.fn.attr()
- Book of Evil 树双向DFS
- 设置picturebox 背景为透明
- u- boot 的认识
- Google Protocol Buffer 的使用和原理
- 如何在ArcGIS Viewer for Silverlight中使用WCF服务完成降雨量专题图显示(二)
- MSM--Memcached_Session_Manager介绍及使用
- 【WindowsXP自带的压缩功能使用详解】
- JS 根据表格(TABLE)模板增加一行(ROW)
- 正则表达式